Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

int main()
{
  int n;
  cin>>n;
  vector<double> dp(n+1),ip(n);
  //dp[i]->i heads
  for(double &x: ip)
  cin>>x;

  dp[0]=1;
  for(int i=0;i<n;i++)
  {
    for(int j=i+1;j>=0;j--)
    {
      dp[j]=dp[j]*(1-ip[i])+(j==0?0:dp[j-1]*ip[i]);
    }
  }
  double sum=0;
  for(int i=0;i<=n;i++)
  {
    int t=n-i;
    if(i>t)
    sum+=dp[i];
  }
  printf("%.10lf\n",sum);

  return 0;
}